Question

An airplane is flying through a thundercloud at a height of 2010 m. (A very dangerous thing to do because of updrafts, turbulence, and the possibility of electric discharge.) If there is a charge concentration of 46.9 C at height 4090 m within the cloud and −59.7 C at height 570 m, what is the magnitude of the electric field E at the aircraft? The Coulomb constant is 8.98755 × 109 N · m2 /C 2 . Answer in units of V/m.

Answer 1

The value is
`E = 3.5619 *10^(5) V/m`

Step-by-step explanation:

From the question we are told that

The height of the airplane is
`h = 2010 \ m`

The first charge concentration is
`Q = 46.9 \ C`

The height of the first charge concentration is
`h_1 = 4090 \ m`

The second charge concentration is
`Q_2 = −59.7 \ C`

The height of the second charge concentration is
`h_1 = 570 \ m`

The electric field due to the first charge concentration is

`E_1 = (k * |Q_1|)/((h_1 -h)^2)`

Here k is the Coulomb constant given in the question

`E_1 = (8.98755*10^9 * |46.9|)/((4090 - 2010)^2)`

`E_1 = 9.7429 *10^(4) \ V/m`

The electric field due to the second charge concentration is

`E_2 = (k * |-59.7|)/((h -h_2)^2)`

`E_2 = (8.98755*10^9 * |-59.7|)/((2010 - 570)^2)`

`E_2 = 2.5876 *10^(5) V/m`

Generally the superposition principle can be applied in this question as follows

`E = E_1 + E_2`

=>
`E = 9.7429 *10^(4)+ 2.5876 *10^(5)`

=>
`E = 3.5619 *10^(5) V/m`